Absolute Value of Measurable Function is Measurable

Theorem

Let $\struct {X, \Sigma}$ be a measurable space.

Let $f : X \to \overline \R$ be a $\Sigma$-measurable function.

Then:

$\size f$ is a $\Sigma$-measurable function.

Proof

From Characterization of Measurable Functions, it suffices to show that for each real number $t \in \R$, we have:

$\set {x \in X : \size {\map f x} \le t} \in \Sigma$

If $t < 0$, we have:

$\set {x \in X : \size {\map f x} \le t} = \O$

So, from Properties of Algebras of Sets, we have:

$\set {x \in X : \size {\map f x} \le t} \in \Sigma$

if $t < 0$.

If $t \ge 0$, we can write:

\(\ds \set {x \in X : \size {\map f x} \le t}\)

\(=\)

\(\ds \set {x \in X : -t \le \map f x \le t}\)

\(\ds \)

\(=\)

\(\ds \set {x \in X : -t \le \map f x} \cap \set {x \in X : \map f x \le t}\)

Since $f$ is $\Sigma$-measurable, we have that both:

$\set {x \in X : -t \le \map f x} \in \Sigma$

and:

$\set {x \in X : \map f x \le t} \in \Sigma$

from Characterization of Measurable Functions.

From Properties of Algebras of Sets, the intersection of any two sets in $\Sigma$ is contained in $\Sigma$.

So:

$\set {x \in X : -t \le \map f x} \cap \set {x \in X : \map f x \le t} \in \Sigma$

if $t \ge 0$.

That is:

$\set {x \in X : \size {\map f x} \le t} \in \Sigma$

if $t \ge 0$.

So:

$\set {x \in X : \size {\map f x} \le t} \in \Sigma$

for all $t \in \R$.

$\blacksquare$